smc

Let $\frac{y}{x}=k$, so that $y=kx$. Substituting this into the given equation $(x-3{)}^2+(y-3{)}^2=6$ yields $(x-3{)}^2+(kx-3{)}^2=6$. Multiplying this out and forming it into a quadratic yields $(1+k^2)x^2+(-6-6k)x+12=0$. We want $x$ to be a real number, so we must have the discriminant $\ge 0$. The discriminant is $(-6-6k{)}^2-4(1+k^2)(12)=36k^2+72k+36-48-48k^2=-12k^2+72k-12$. Therefore, we must have $-12k^2+72k-12\ge 0$, or $k^2-6k+1\le 0$. The roots of this quadratic, using the quadratic formula, are $3\pm 2\sqrt{2}$, so the quadratic can be factored as $(k-(3-2\sqrt{2}))(k-(3+2\sqrt{2}))\le 0$. We can now separate this into $3$ cases: Case 1: $k<3-2\sqrt{2}$ Then, both terms in the factored quadratic are negative, so the inequality doesn't hold. Case 2: $3-2\sqrt{2}<k<3+2\sqrt{2}$ Then, the first term is positive and the second is negative and the second is positive, so the inequality holds. Case 3: $k>3+2\sqrt{2}$ Then, both terms are positive, so the inequality doesn't hold. Also, when $k=3-2\sqrt{2}$ or $k=3+2\sqrt{2}$, the equality holds. Therefore, we must have $3-2\sqrt{2}\le k\le 3+2\sqrt{2}$, and the maximum value of $k=\frac{y}{x}$ is $3+2\sqrt{2},\boxed{3+2\sqrt{2}}$.